If y=y(x) is the solution of the differential equation \fracd y~d x+2 y=\sin (2 x), y(0)=\frac34 , then y(\frac\pi8) is equal to :

Differential Equations PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q50 with Solution

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Correct answer: C

$\frac{dy}{dx}+2y=\sin(2x), \qquad y(0)=\frac34$

This is a linear first-order differential equation.

For $\frac{dy}{dx}+P(x)y=Q(x),$ the integrating factor is $\text{I.F.}=e^{\int P(x)\,dx}=e^{\int 2\,dx}=e^{2x}.$

$e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}\sin(2x).$

The left side becomes:

$\frac{d}{dx}\left(ye^{2x}\right)=e^{2x}\sin(2x).$

So,

$ye^{2x}=\int e^{2x}\sin(2x)\,dx + C.$

Use the standard result:

$\int e^{ax}\sin(bx)\,dx=\frac{e^{ax}}{a^2+b^2}\left(a\sin bx-b\cos bx\right).$

Here $a=2$ , $b=2$ , so

=\frac{e^{2x}}{8}(2\sin 2x-2\cos 2x).$$ Thus, $$\int e^{2x}\sin(2x)\,dx=\frac{e^{2x}}{4}(\sin 2x-\cos 2x).$$ Hence, $$ye^{2x}=\frac{e^{2x}}{4}(\sin 2x-\cos 2x)+C.$$ Dividing by $e^{2x}$, $$y=\frac14(\sin 2x-\cos 2x)+Ce^{-2x}.$$ 5. **Use the initial condition** Given $y(0)=\frac34$: $$\frac34=\frac14(\sin 0-\cos 0)+C e^0$$ $$\frac34=\frac14(0-1)+C$$ $$\frac34=-\frac14+C$$ $$C=1.$$ So the solution is $$y=\frac14(\sin 2x-\cos 2x)+e^{-2x}.$$ 6. **Find $y\left(\frac\pi8\right)$** Substitute $x=\frac\pi8$: $$2x=\frac\pi4$$ and $$\sin\frac\pi4=\cos\frac\pi4=\frac{1}{\sqrt2}.$$ Therefore, $$\frac14\left(\sin\frac\pi4-\cos\frac\pi4\right)=\frac14\left(\frac{1}{\sqrt2}-\frac{1}{\sqrt2}\right)=0.$$ So, $$y\left(\frac\pi8\right)=e^{-2\cdot \pi/8}=e^{-\pi/4}.$$ 7. **Compare with options** $$y\left(\frac\pi8\right)=e^{-\pi/4}$$ This matches **Option C**.